Question

Решите уравнения: a) sin 3x = 1 б) 1-4 sinx cos x = 0 B) COS X cos3x = cos 2x cos 4x ​

Answer 1

$sin3x=1\\3x=\frac{\pi }{2} +2\pi n\\x=\frac{\pi }{6}+\frac{2\pi }{3} n$

$1-4sinxcosx=0\\1-2sin2x=0\\sin2x=\frac{1}{2} \\2x_{1} =\frac{\pi }{6} +2\pi n\\x_{1} =\frac{\pi }{12} +\pi n\\2x_{2} =\frac{5\pi }{6} +2\pi k\\x_{2} =\frac{5\pi }{12} +\pi k$

$cosx-cos3x=cos2x-cos4x\\sin^{2} x(2cosx+4sin^{2} x-3)=0\\sin^{2} x=0\\x_{1} =\pi k\\-4cos^{2}x+2cosx+1=0\\ cosx=\frac{1+\sqrt{5} }{4} \\cosx=\frac{1+\sqrt{5} }{4} \\x_{2} =arccos(\frac{1+\sqrt{5} }{4} )+2\pi n\\x_{3} =-arccos(\frac{1+\sqrt{5} }{4} )+2\pi z\\x_{4} =arccos(\frac{1-\sqrt{5} }{4} )+2\pi w\\x_{5} =-arccos(\frac{1-\sqrt{5} }{4} )+2\pi j$